CODEWITHSUNDEEP
mathlogicbit-manipulationbitwise-operators
Fit Dev
Fit Dev

Reputation: 3689

Is there a way to use Bitwise Operators instead of multiple Ternary Operators?

Is there a way to get a result of applying a ternary ?: operator to individual bits without actually computing it one bit at a time, i.e. using some combination of and-s, or-s, xor-s, and not-s?

In other words, given we have 3 ints (regardless of bit length): a, b, and mask m, the result r should basically be: r = m ? a : b, except on a per-bit basis.

Assuming a = 0b0101; b = 0b1110; and m = 0b0110, the result r should be 0b1100. So, if bit n in mask m is 1 (true), use the n-th bit from a; otherwise (if bit n in mask m is 0 (false)) then use the n-th bit from b.

Upvotes: 1

Views: 237

Answers (2)

Fit Dev
Fit Dev

Reputation: 3689

For explanatory purposes, I soon realized that the problem can be thought of as taking all valid 1s from a and b by using the mask m and combining the results. Then the problem becomes trivial, as illustrated in the truth table:

-------------------------------------------
         a:   0   0   0   0   1   1   1   1
         b:   1   1   0   0   1   1   0   0
         m:   1   0   1   0   1   0   1   0   // 1 = the positions from a; 0 = positions from b

        ~m:   0   1   0   1   0   1   0   1
 
 x = m & a:   0   0   0   0   1   0   1   0   // take all valid 1s from a
y = ~m & b:   0   1   0   0   0   1   0   0   // take all valid 1s from b

 r = x | y:   0   1   0   0   1   1   1   0   // combine: r = (m & a) | (~m & b)
-------------------------------------------
  EXPECTED:   0   1   0   0   1   1   1   0

An alternative implementation using xor-s: r = b ^ (m & (a ^ b))

-------------------------------------------
         a:   0   0   0   0   1   1   1   1
         b:   1   1   0   0   1   1   0   0
         m:   1   0   1   0   1   0   1   0   // 1 = the positions from a; 0 = positions from b

 u = a ^ b:   1   1   0   0   0   0   1   1
 v = m & u:   1   0   0   0   0   0   1   0

 r = b ^ v:   0   1   0   0   1   1   1   0
-------------------------------------------
  EXPECTED:   0   1   0   0   1   1   1   0

Upvotes: 0

Stef
Stef

Reputation: 15525

If m, a and b are all guaranteed to be 0 or 1, then m ? a : b is equivalent to (m && a) || (!m && b). You can see that by injecting m=0 or m=1 in the formula and simplifying it.

This works bitwise too; bitwise not is usually noted ~; so the "bitwise ternary operator" that you are looking for can be expressed as:

(m&a)|((~m)&b)

Upvotes: 1

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